1083 11.5 Mathematical Induction EXAMPLE 2 Proving an Inequality Statement Prove that if x is a real number strictly between 0 and 1, then for every positive integer n, it follows that 0 6xn 61. SOLUTION Step 1 Let Sn represent the given statement. Here S1 is the statement if 0 6x 61, then 0 6x1 61, which is true. Step 2 Sk is the statement if 0 6x 61, then 0 6xk 61. To show that this implies that Sk+1 is true, multiply each of the three parts of 0 6xk 61 by x. 0 6 xk 61 x # 0 6x # xk 6x # 1 Use the fact that 0 6x. 0 6 xk+1 6x Simplify. We now use a technique that allows us to reach our desired goal. From Step 1 we know that x 61, so in the inequality 0 6xk+1 6x, we can replace x with any greater value and the inequality is preserved. Because 1 is greater than x, replace x with 1. 0 6xk+1 61 This is the statement Sk+1. This work shows that if Sk is true, then Sk+1 is true. Therefore, the given statement Sn is true for every positive integer n. S Now Try Exercise 21. Generalized Principle of Mathematical Induction Some statements Sn are not true for the first few values of n but are true for all values of n that are greater than or equal to some fixed integer j. The following generalized form of the principle of mathematical induction covers these cases. Generalized Principle of Mathematical Induction Let Sn be a statement concerning the positive integer n. Let j be a fixed positive integer. Suppose that both of the following are satisfied. Step 1 Sj is true. Step 2 For any positive integer k, k Ú j, Sk implies Sk+1. Then Sn is true for all positive integers n, where n Ú j. CAUTION Notice that the left side of the statement Sn in Example 1 includes all the terms up to the nth term, as well as the nth term.
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